Answer
$$f'\left( w \right) = \frac{{\cos \left( {{{\sec }^{ - 1}}2w} \right)}}{{\left| w \right|\sqrt {4{w^2} - 1} }}$$
Work Step by Step
$$\eqalign{
& f\left( w \right) = \sin \left( {{{\sec }^{ - 1}}2w} \right) \cr
& {\text{find the derivative}}{\text{, differentiate both sides with respect to }}w \cr
& f'\left( w \right) = \frac{d}{{dw}}\left[ {\sin \left( {{{\sec }^{ - 1}}2w} \right)} \right] \cr
& {\text{use }}\frac{d}{{dw}}\left[ {\sin u} \right] = \cos u\frac{{du}}{{dw}}.{\text{ let }}u = {\sec ^{ - 1}}2w \cr
& {\text{then}} \cr
& f'\left( w \right) = \cos \left( {{{\sec }^{ - 1}}2w} \right)\frac{d}{{dw}}\left[ {{{\sec }^{ - 1}}2w} \right] \cr
& {\text{use }}\frac{d}{{dw}}\left[ {{{\sec }^{ - 1}}u} \right] = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dw}}.{\text{ let }}u = 2w \cr
& {\text{then}} \cr
& f'\left( w \right) = \cos \left( {{{\sec }^{ - 1}}2w} \right)\frac{1}{{\left| {2w} \right|\sqrt {{{\left( {2w} \right)}^2} - 1} }}\frac{d}{{dw}}\left[ {2w} \right] \cr
& f'\left( w \right) = \cos \left( {{{\sec }^{ - 1}}2w} \right)\frac{1}{{\left| {2w} \right|\sqrt {{{\left( {2w} \right)}^2} - 1} }}\left( 2 \right) \cr
& {\text{simplifying}} \cr
& f'\left( w \right) = \cos \left( {{{\sec }^{ - 1}}2w} \right)\frac{1}{{2\left| w \right|\sqrt {4{w^2} - 1} }}\left( 2 \right) \cr
& f'\left( w \right) = \frac{{\cos \left( {{{\sec }^{ - 1}}2w} \right)}}{{\left| w \right|\sqrt {4{w^2} - 1} }} \cr} $$
